Question

find the sum of an AP 13+19+25.....to 10 terms

Answer 1

Step-by-step explanation:

d=19-13=6

n=10

a=13

s10=n/2[2a+(n-1)d]

s10=5[26+(54)]

s10=5×80

s10=400

Answer 2

Given:-

The term in A.P

=> 13+19+25+....... .....10 terms

To find :-

The sum of given terms

Now

AT.Q

The formula

$\boxed{\sf{S_n=\dfrac{n}{2}(2a+(n-1)d}}$

where n is the number of terms

d is the common difference

and a is 1st term

Now

• we have

$\sf\bullet a=13\\ \\ \sf\bullet d=a_2-a_1\\ \sf\bullet d= 19-13 =6\\ \\ \sf \bullet n=10$

• Now find the sum

$\sf\implies S_{10}=\dfrac{10}{2}[2\times 13+(10-1)6]\\ \\ \sf\implies S_{10}= \dfrac{10}{2}(26+9\times 6)\\ \\ \sf\implies S_{10}=\cancel{\dfrac{10}{2}}(26+54)\\ \\ \sf\implies S_{10}=5\times 80\\ \\ \sf\implies S_{10}=400$

So the sum is 400